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junior college 1 | H1 Maths
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Cant seem to figure this out since u does not = to c x v meaning the lines are not parallel as well...
Date Posted: 4 years ago
Views: 302
Could be like this :
L1 : r = (1 2 -1) + λ(2 -4 1)
= (1+2λ 2-4λ λ-1)
L2 : r = (-2 1 1) + μ(1 5 α)
= (μ-2 1+5μ 1+αμ)
For them to intersect,
1 + 2λ = μ - 2 ①
2 - 4λ = 1 + 5μ ②
λ - 1 = 1 + αμ ③
From ①,
μ = 3 + 2λ
Sub this into ②,
2 - 4λ = 1 + 5(3 + 2λ)
2 - 4λ = 16 + 10λ
14λ = -14
λ = -1
Sub λ = -1 into ①,
μ = 3 + 2(-1) = 1
Sub μ = 1, λ = -1 into ③,
-1 - 1 = 1 + α(1)
-2 = 1 + α
α = -3
So as long as a ≠ -3, the lines will not intersect
L1 : r = (1 2 -1) + λ(2 -4 1)
= (1+2λ 2-4λ λ-1)
L2 : r = (-2 1 1) + μ(1 5 α)
= (μ-2 1+5μ 1+αμ)
For them to intersect,
1 + 2λ = μ - 2 ①
2 - 4λ = 1 + 5μ ②
λ - 1 = 1 + αμ ③
From ①,
μ = 3 + 2λ
Sub this into ②,
2 - 4λ = 1 + 5(3 + 2λ)
2 - 4λ = 16 + 10λ
14λ = -14
λ = -1
Sub λ = -1 into ①,
μ = 3 + 2(-1) = 1
Sub μ = 1, λ = -1 into ③,
-1 - 1 = 1 + α(1)
-2 = 1 + α
α = -3
So as long as a ≠ -3, the lines will not intersect
Hmmm okay i got it thanks
See 1 Answer
done
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Pls verify...
Date Posted:
4 years ago
So basically there are no possible solutions at all for a then..
The solution is "alpha is an element of Real number, except -3". Just that this alpha is not one value only, as what the question asks "all possible values of alpha"...
Thankyou!
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